QuestionMay 23, 2025

a. When 8.55 g diethyl ether (C_(4)H_(10)O) is dissolved in 100 g of benzene, what will be the boiling point of this solution? K_(b) for benzene is 2.53^circ C/m and its boiling point is 81.10^circ C square [ Select ] v b. The freezing point of a solution of barium phosphate where 395 g barium phosphate is dissolved in 888 g of di water is [Select] ^circ C.K_(f)=-1.86(^circ C)/(m)

a. When 8.55 g diethyl ether (C_(4)H_(10)O) is dissolved in 100 g of benzene, what will be the boiling point of this solution? K_(b) for benzene is 2.53^circ C/m and its boiling point is 81.10^circ C square [ Select ] v b. The freezing point of a solution of barium phosphate where 395 g barium phosphate is dissolved in 888 g of di water is [Select] ^circ C.K_(f)=-1.86(^circ C)/(m)
a. When 8.55 g diethyl ether (C_(4)H_(10)O) is dissolved in 100 g of benzene, what will be the boiling
point of this solution? K_(b) for benzene is 2.53^circ C/m and its boiling point is 81.10^circ C
square  [ Select ]
v
b. The freezing point of a solution of barium phosphate where 395 g barium phosphate is
dissolved in 888 g of di water is [Select] ^circ C.K_(f)=-1.86(^circ C)/(m)

Solution
4.4(259 votes)

Answer

a. 84.017^{\circ}C ### b. -6.918^{\circ}C Explanation 1. Calculate molality of diethyl ether in benzene Molar mass of C_{4}H_{10}O is 74.12 \, g/mol. Molality m = \frac{8.55 \, g}{74.12 \, g/mol} \times \frac{1000}{100 \, g} = 1.153 \, m. 2. Calculate boiling point elevation Boiling point elevation \Delta T_b = K_b \times m = 2.53^{\circ}C/m \times 1.153 \, m = 2.917^{\circ}C. 3. Determine new boiling point New boiling point = 81.10^{\circ}C + 2.917^{\circ}C = 84.017^{\circ}C. 4. Calculate molality of barium phosphate in water Molar mass of Ba_3(PO_4)_2 is 601.93 \, g/mol. Molality m = \frac{395 \, g}{601.93 \, g/mol} \times \frac{1000}{888 \, g} = 0.744 \, m. 5. Calculate freezing point depression Freezing point depression \Delta T_f = i \times K_f \times m = 5 \times (-1.86^{\circ}C/m) \times 0.744 \, m = -6.918^{\circ}C (i = 5 for Ba_3(PO_4)_2). 6. Determine new freezing point New freezing point = 0^{\circ}C - 6.918^{\circ}C = -6.918^{\circ}C.

Explanation

1. Calculate molality of diethyl ether in benzene<br /> Molar mass of $C_{4}H_{10}O$ is $74.12 \, g/mol$. Molality $m = \frac{8.55 \, g}{74.12 \, g/mol} \times \frac{1000}{100 \, g} = 1.153 \, m$.<br /><br />2. Calculate boiling point elevation<br /> Boiling point elevation $\Delta T_b = K_b \times m = 2.53^{\circ}C/m \times 1.153 \, m = 2.917^{\circ}C$.<br /><br />3. Determine new boiling point<br /> New boiling point $= 81.10^{\circ}C + 2.917^{\circ}C = 84.017^{\circ}C$.<br /><br />4. Calculate molality of barium phosphate in water<br /> Molar mass of $Ba_3(PO_4)_2$ is $601.93 \, g/mol$. Molality $m = \frac{395 \, g}{601.93 \, g/mol} \times \frac{1000}{888 \, g} = 0.744 \, m$.<br /><br />5. Calculate freezing point depression<br /> Freezing point depression $\Delta T_f = i \times K_f \times m = 5 \times (-1.86^{\circ}C/m) \times 0.744 \, m = -6.918^{\circ}C$ (i = 5 for $Ba_3(PO_4)_2$).<br /><br />6. Determine new freezing point<br /> New freezing point $= 0^{\circ}C - 6.918^{\circ}C = -6.918^{\circ}C$.
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